Re: Question

In article <1145996666.531448.45830@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"zuhair" <zaljohar@xxxxxxxxx> wrote:

Virgil wrote:
In article <1145991892.167851.295010@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"zuhair" <zaljohar@xxxxxxxxx> wrote:


Second you think that number 1 can be represented as 0.99999........ (
decimal) , well I think that this is only an approximate representation
that is not correct.
The only representation of 1 is 1.0000......... and to me
0.9999........... < 1.

If 0.9999........... < 1 then x = 1 - 0.9999... > 0

So how much larger than zero is this x?

Is it large enough so that x/2 is smaller?

[x/n] + [x/(n^2)] + [x/(n^3)] +.............+ [ x/(n^k)] = [x/(n-1)] -
[ x/{ ( n^k) ( n-1 ) } ]

For n=2,3,4,5,.............
and K= 1,2,3,4,5,............

now 0.99999............ = 9/ 10 + 9 / 10^2 + 9 / 10^3 +
....................= 9/9 - 9/[ (10^Omega) *9} = 1 - [1/(10^Omega)]

Now [1/(10^Omega)] > zero

because if [1/(10^Omega)] = 0 then 0* 10^Omega = 1 which is
impossible.

Then 0.9999......... < 1

Zuhair

Does not answer the question of whether x/2 is smaller than x.

and is a false summation formula anyway.

The infinite sum 1 + a + a^2 + a^3 + ... is undefined for |a| >=1 and
converges to the value 1/(1-a) when |a| < 1.
so that (9/10)*(1 + 1/10 + 1/100 + ...) = (9/10)(1/(1 - 1/10)) = 1
.

Relevant Pages

  • Re: Question
    ... So how much larger than zero is this x? ... Does not answer the question of whether x/2 is smaller than x. ... and is a false summation formula anyway. ... The summation formula I have provided is the true one. ...
    (sci.math)
  • Re: Question
    ... Virgil wrote: ... well I think that this is only an approximate representation ... So how much larger than zero is this x? ... Is it large enough so that x/2 is smaller? ...
    (sci.math)
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